Problem 12 - Applied Mathematics

Question

A cylinder of weight } 100 \mathrm{~N} \text { rests in the angle between a smooth vertical wall }and a smooth plane inclined at \(30^{\circ}\) to the wall. Find the thrusts of the cylinder on the wall and the plane.

Step-by-Step Solution

Answer

The thrust on the wall is 57.75 N and on the plane is 115.5 N.

Step 1: Draw a Diagram and Identify Forces

Draw a diagram of the situation. A cylinder rests between a vertical wall and an inclined plane at an angle of \(30^{\text{°}}\). Identify the forces acting on the cylinder: 1) the weight (\(W = 100 \text{ N}\)) acting vertically downward, 2) the normal force (\(N_1\)) from the wall acting horizontally, and 3) the normal force (\(N_2\)) from the inclined plane acting perpendicular to the plane.

Step 2: Resolve the Weight into Components

Resolve the weight (\(W = 100 \text{ N}\)) into components parallel and perpendicular to the inclined plane. Because the angle between the weight and the inclined plane is \(30^{\text{°}}\), use trigonometric functions: \[ W_{parallel} = W \sin{30^{\text{°}}} = 100 \sin{30^{\text{°}}} = 50 \text{ N} \] \[ W_{perpendicular} = W \cos{30^{\text{°}}} = 100 \cos{30^{\text{°}}} = 86.6 \text{ N} \]

Step 3: Apply Equilibrium Conditions

For the cylinder to be in equilibrium, the sum of forces in the horizontal and vertical directions must each be zero. Consider the following equilibrium equations: Horizontal: \[ N_1 = N_2 \sin 30^{\text{°}} \] Vertical: \[ N_2 \cos 30^{\text{°}} = W \]

Step 4: Solve for the Normal Forces

Use the vertical equilibrium equation to solve for \(N_2\): \[ N_2 \cos 30^{\text{°}} = 100 \] \[ N_2 \left( \frac{\root{3}/2} \right) = 100 \] \[ N_2 = \frac{100}{\cos 30^{\text{°}}} = \frac{100}{\root{3}/2} = \frac{200}{\root{3}} \approx 115.5 \text{ N} \] Now use the horizontal equilibrium equation to find \(N_1\): \[ N_1 = N_2 \sin 30^{\text{°}} = 115.5 \times \frac{1}{2} = 57.75 \text{ N} \]

Key Concepts

Normal Force

Normal force is a fundamental concept in physics that you need to understand to solve equilibrium problems. This force is perpendicular to the surface with which an object is in contact. For our exercise, there are two normal forces:

• The normal force from the wall (\(N_1\)) acts horizontally.
• The normal force from the inclined plane (\(N_2\)) acts perpendicular to the plane.

These normal forces are critical because they balance out the forces acting in both horizontal and vertical directions, allowing the cylinder to remain in equilibrium.

Resolving Forces

When working with forces, especially in equilibrium conditions, you'll often need to 'resolve' a force into components. This means breaking it down into parts that act in different directions. In our exercise, we resolve the weight (\(W\)) of the cylinder into two components:

• Parallel to the inclined plane (\(W_{parallel}\)): \(W \sin{30^{\text{°}}}\), which turns out to be 50 N.
• Perpendicular to the inclined plane (\(W_{perpendicular}\)): \(W \cos{30^{\text{°}}}\), which turns out to be 86.6 N.

These components help us understand how the weight of the cylinder interacts with the inclined plane and the wall, making it easier to apply equilibrium conditions.

Trigonometric Functions

Trigonometric functions are indispensable for resolving forces and understanding their components. In this exercise, we use sine and cosine functions to find how much of the cylinder's weight acts parallel and perpendicular to the inclined plane. Here's a quick reminder of the key functions we use:

• The sine function (\(\text{sin}\)): \( \text{sin}\left(\theta\right) = \frac{\text{opposite}}{\text{hypotenuse}}\)
• The cosine function (\(\text{cos}\)): \(\ text {cos}\left(\theta\right) = \frac{\text{adjacent}}{\text{hypotenuse}}\)

In our problem, the angle between the weight and the inclined plane is 30°.Using \( \text{sin}\left( 30^{\text{°}} \right) \) and \( \cos\left( 30^{\text{°}} \right) \) lets us easily divide the weight into components (\( cos: 86.6 \text{N}\) and \( sin: 50 \text{N}\)).

Newton's Laws of Motion

Newton's laws of motion are at the core of understanding equilibrium conditions. The most relevant one for this exercise is Newton's first law: an object at rest stays at rest unless acted upon by a net external force. For the cylinder to be in equilibrium:

• The sum of all horizontal forces must be zero.
• The sum of all vertical forces must be zero.

In the exercise, we utilize these principles: \( N_1 = N_2 \sin{30^{\text{°}}} \) for horizontal equilibrium, and \( N_2 \cos{30^{\text{°}}} = W\) for vertical equilibrium. This means: \( N_1\) and \(N_2\) balance out the cylinder's weight, ensuring it stays put.